Question: Divide the following complex numbers. $ \dfrac{-10+2i}{2+2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2-2i}$ $ \dfrac{-10+2i}{2+2i} = \dfrac{-10+2i}{2+2i} \cdot \dfrac{{2-2i}}{{2-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-10+2i) \cdot (2-2i)} {(2+2i) \cdot (2-2i)} = \dfrac{(-10+2i) \cdot (2-2i)} {2^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-10+2i) \cdot (2-2i)} {(2)^2 - (2i)^2} = $ $ \dfrac{(-10+2i) \cdot (2-2i)} {4 + 4} = $ $ \dfrac{(-10+2i) \cdot (2-2i)} {8} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-10+2i}) \cdot ({2-2i})} {8} = $ $ \dfrac{{-10} \cdot {2} + {2} \cdot {2 i} + {-10} \cdot {-2 i} + {2} \cdot {-2 i^2}} {8} $ Evaluate each product of two numbers. $ \dfrac{-20 + 4i + 20i - 4 i^2} {8} $ Finally, simplify the fraction. $ \dfrac{-20 + 4i + 20i + 4} {8} = \dfrac{-16 + 24i} {8} = -2+3i $